The sum of five consecutive multiples of four is 200. Find the smallest multiple.

Correct option is D

Let the five consecutive multiples of four be 4a, 4b, 4c, 4d, 4e;
where a=x, b=x+1, c=x+2, d=x+3 and e=x+4 ………(I)
ATQ,
4a+4b+4c+4d+4e=200
a+b+c+d+e=50
Or, x+x+1+x+2+x+3+x+4=50 ……..from (I)
5x=50-10
x=8
So, smallest multiple=4a=4x=4×8=32